For ps in the example above r2 without ps 9861 r2

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For PS in the example above, R2 without PS = .9861, R2 with PS = .9907 (.9907 - .9861) / (1 - .9861) = .3309 3.922 / (3.922 + (36-5)) = .3314 (rounding) ™     13/33
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Part 5: Regression Algebra and Fit Comparing fits of regressions Make sure the denominator in R2 is the same - i.e., same left hand side variable. Example, linear vs. loglinear. Loglinear will almost always appear to fit better because taking logs reduces variation. ™    14/33
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Part 5: Regression Algebra and Fit ™    15/33
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Part 5: Regression Algebra and Fit (Linearly) Transformed Data p How does linear transformation affect the results of least squares? Z = XP for KxK nonsingular P p Based on X , b = ( XX )-1 X’y . p You can show (just multiply it out), the coefficients when y is regressed on Z are c = P -1 b p “Fitted value” is Zc = XPP -1 b = Xb . The same!! p Residuals from using Z are y - Zc = y - Xb (we just proved this.). The same!! p Sum of squared residuals must be identical, as y - Xb = e = y - Zc . p R2 must also be identical, as R2 = 1 - ee / y’M 0 y (!!). ™    16/33
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Part 5: Regression Algebra and Fit Linear Transformation p Xb is the projection of y into the column space of X . Zc is the projection of y into the column space of Z . But, since the columns of Z are just linear combinations of those of X , the column space of Z must be identical to that of X . Therefore, the projection of y into the former must be the same as the latter, which now produces the other results.) p What are the practical implications of this result? n Transformation does not affect the fit of a model to a body of data. n Transformation does affect the “estimates.” If b is an estimate of something ( ), then c cannot be an estimate of - it must be an estimate of P -1 , which might have no meaning at all. ™    17/33
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Part 5: Regression Algebra and Fit Principal Components p Z = XC n Fewer columns than X n Includes as much ‘variation’ of X as possible n Columns of Z are orthogonal p Why do we do this? n Collinearity n Combine variables of ambiguous identity such as test scores as measures of ‘ability’ p How do we do this? Later in the course. Requires some further results from matrix algebra. ™    18/33
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Part 5: Regression Algebra and Fit What is a Principal Component? p X = a data matrix (deviations from means) p z = Xp = linear combination of the columns of X . p Choose p to maximize the variation of z . p How? p = eigenvector that corresponds to the largest eigenvalue of X’X ™    19/33
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Part 5: Regression Algebra and Fit +----------------------------------------------------+ | Movie Regression. Opening Week Box for 62 Films | | Ordinary least squares regression | | LHS=LOGBOX Mean = 16.47993 | | Standard deviation = .9429722 | | Number of observs. = 62 | | Residuals Sum of squares = 20.54972 | | Standard error of e = .6475971 | | Fit R-squared = .6211405 | | Adjusted R-squared = .5283586 | +----------------------------------------------------+ +--------+--------------+----------------+--------+--------+----------+ |Variable| Coefficient | Standard Error |t-ratio |P[|T|>t]| Mean of X| +--------+--------------+----------------+--------+--------+----------+ |Constant| 12.5388*** .98766 12.695 .0000 | |LOGBUDGT| .23193 .18346 1.264 .2122 3.71468| |STARPOWR| .00175 .01303 .135 .8935 18.0316| |SEQUEL | .43480 .29668 1.466 .1492 .14516| |MPRATING| -.26265* .14179 -1.852 .0700 2.96774| |ACTION | -.83091*** .29297 -2.836 .0066 .22581| |COMEDY | -.03344 .23626 -.142 .8880 .32258| |ANIMATED| -.82655** .38407 -2.152 .0363 .09677| |HORROR | .33094 .36318 .911 .3666 .09677| 4 INTERNET BUZZ VARIABLES |LOGADCT | .29451** .13146 2.240 .0296 8.16947| |LOGCMSON| .05950 .12633 .471 .6397 3.60648| |LOGFNDGO| .02322 .11460 .203 .8403 5.95764| |CNTWAIT3| 2.59489*** .90981 2.852 .0063 .48242| +--------+------------------------------------------------------------+ ™    20/33
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